There are models of ZF in which some Cantor cubes are non-compact. If a Cantor cube $X= 2^J$ is non-compact, there does not exist a compactification $\alpha X$ of $X$ such that, for each $j\in J$, the projection $\pi_j: X\to {0, 1}$ is continuously extendable. Suppose that M is a model of ZF in which a Cantor cube $X=2^J$ is not compact. Then, of course, $X$ is not locally compact, so it might be hard to find a Hausdorff compactification of $X$ in M. I wonder whether there is any Hausdorff compactification of $X$ in $M$. Perhaps, some researchers know a satisfactory answer to my question and can show to me a proper piece of literature about it. I would be grateful for it. I have prepared a talk at a conference about ZF-theory of Hausdorff compactifications, some constructions of strange Hausdorff compactifications that are not cube-compact of some Tychonoff spaces are included in it.
product of completely regular space is completely regular, so why not have a Hausdorff compactification.
Subrata, to prove that a completely regular space has a Hausdorff compactification you need of Tychonoff theorem (any product of compact spaces is compact), and it is equivalent to the Axiom of Choice. Eliza's question was made in the context of ZF, that is, Axiom of Choice is out of the picture.
Thank you for your comments. Dear Subrata, simply, in some ZF-models without AC, some Tychonoff cubes can be non-compact and even compact Hausdorff spaces can fail to be completely regular, I recommend to you the book "Axiom of Choice" by Horst Herrlich. I've already prepared my talk for a conference on November 18-20 about ZF-theory of Hausdorff compactifications. Many theorems of ZFC-theory of Hausdorff compactifications simply fail in some models of ZF. Some theorems can be preserved for those compactifications that are Tychonoff spaces. To construct some compactifications, especially of locally compact Hausdorff spaces, only ZF is enough. However, when a Cantor cube is non-compact, it is not locally compact and those constructions cannot be applied. Eliza
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